Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 3x}{x + 3} = \dfrac{-12x - 18}{x + 3}$
Explanation: Multiply both sides by $x + 3$ $ \dfrac{x^2 - 3x}{x + 3} (x + 3) = \dfrac{-12x - 18}{x + 3} (x + 3)$ $ x^2 - 3x = -12x - 18$ Subtract $-12x - 18$ from both sides: $ x^2 - 3x - (-12x - 18) = -12x - 18 - (-12x - 18)$ $ x^2 - 3x + 12x + 18 = 0$ $ x^2 + 9x + 18 = 0$ Factor the expression: $ (x + 6)(x + 3) = 0$ Therefore $x = -6$ or $x = -3$ At $x = -3$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -3$, it is an extraneous solution.